Answer
$0$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{1-\cos x}{\tan x+x (\sec^2 x)}=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{\sin x}{2 \sec^2 x+2x\sec^2 x \tan x}=\dfrac{0}{2+0}=0$